[tex]a=\log_{45}3=\dfrac{1}{\log_345}=\dfrac{1}{\log_39+\log_35}=\dfrac{1}{2+\log_35}\\\\\\a=\dfrac{1}{2+\log_35}\\\\2+\log_35=\dfrac{1}{a}\\\\\log_35=\dfrac{1}{a}-2\\\\\log_35=\dfrac{1}{a}-\dfrac{2a}{a}\\\\\log_35=\dfrac{1-2a}{a}[/tex]
